1

Within QGIS I'm working on a data set that contains species at 100m resolution (eg OSGB 6 fig) and 10m or finer resolution (eg OSGB 8 or 10fig). The objective is to count the number of unique species within a polygon layer and associated buffer layer.

In acknowledging the resolution differences between the lower (6 fig) and higher (8-10fig) resolution, I have created a buffer around the polygons within which any 6fig points are also to be included (ie they fall outside the polygon layer but within the buffer).

While finding unique values for the buffer and the actual polygons is straightforward (count points in polygon), I'm not sure how to combine this for two different operations without inevitably double counting species that lie within both the polygon and the buffer layer. Similarly, counting all points within the buffer does not work as it also includes high resolution outside the polygon within adjacent polygons.

In essence, what I want is a way to combine the count points in polygon operation across two separate but linked features to find the total unique values within both the polygon and buffer layer without having duplicates of the same species included. One possible solution I have thought of is creating a presence/absence matrix for each species within the polygon/buffer layers then combining using field calculator, however I'm struggling to achieve this.

  • 1
    Can you add a sample of your data? It's difficult to understand what you are trying to achieve. – vinh Sep 12 at 14:53
0

If you have two polygon layers which you want to overlay then count points with some unique attribute try:

  1. Intersect the polygon layers
  2. Intersect output (as overlay layer) with points (as input layer)
  3. Then use Statistics by categories to get unique point count per whatever field(s) you choose from the polygon intersection. Or add a field and calculate after step 1 and use this.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.